Let n \geq 5 be an integer. If 9^n-8 n-1=64 \alpha and 6^n-5 n-1=25 \beta, then \alpha-\beta is equal to [29-Jun-2022-Shift-2]

Correct Answer is : nC3(8−5)+nC4(82−52)+.......+nCn(8n−2−5n−2)

1+{ }^n C_2(8-5)+{ }^n C_3(8^2-5^2)+. . . . .+{ }^n C_n(8^{n-1}-5^{n-1})

1+{ }^n C_3(8-5)+{ }^n C_4(8^2-5^2)+. . . . .+{ }^n C_n(8^{n-2}-5^{n-2})

{ }^n C_3(8-5)+{ }^n C_4(8^2-5^2)+. . . . . . .+{ }^n C_n(8^{n-2}-5^{n-2})

{ }^n C_4(8-5)+{ }^n C_5(8^2-5^2)+. . . . . .+{ }^n C_n(8^{n-3}-5^{n-3})

Binomial Theorem Part 2

Section: Mathematics

Question : 94 of 100

Marks: +1, -0

Let n≥5 be an integer. If 9n−8n−1=64α and 6n−5n−1=25β, then α−β is equal to

[29-Jun-2022-Shift-2]

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